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Darina [25.2K]
3 years ago
7

Your friend is 1.2 times taller than you. Your friend is 64.5 inches tall. How tall are you?

Mathematics
2 answers:
vladimir1956 [14]3 years ago
7 0
53.75in is your height.
so this is the equation. x(1.2)=64.5
                                     x=64.5/1.2
                                      x=53.75
Pavel [41]3 years ago
5 0
You are 53.75 inches tall.

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anyanavicka [17]

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x =\dfrac{45 \sqrt{6}}{ 2}

Step-by-step explanation:

From the given information:

The diagrammatic interpretation of what the question is all about can be seen in the diagram attached below.

Now, let V(x) be the time needed for the runner to reach the buoy;

∴ We can say that,

\mathtt{V(x) = \dfrac{70-x}{7}+\dfrac{\sqrt{54^2+x^2}}{5}}

In order to estimate the point along the shore, x meters from B, the runner should  stop running and start swimming if he want to reach the buoy in the least time possible, then we need to differentiate the function of V(x) and relate it to zero.

i.e

The differential of V(x) = V'(x) =0

=\dfrac{d}{dx}\begin {bmatrix} \dfrac{70-x}{7} + \dfrac{\sqrt{54^2+x^2}}{5} \end {bmatrix}= 0

-\dfrac{1}{7}+ \dfrac{1}{5}\times \dfrac{x}{\sqrt{54^2+x^2}}=0

\dfrac{1}{5}\times \dfrac{x}{\sqrt{54^2+x^2}}= \dfrac{1}{7}

\dfrac{5x}{\sqrt{54^2+x^2}}= \dfrac{1}{7}

\dfrac{x}{\sqrt{54^2+x^2}}= \dfrac{1}{\dfrac{7}{5}}

\dfrac{x}{\sqrt{54^2+x^2}}= \dfrac{5}{7}

squaring both sides; we get

\dfrac{x^2}{54^2+x^2}= \dfrac{5^2}{7^2}

\dfrac{x^2}{54^2+x^2}= \dfrac{25}{49}

By cross multiplying; we get

49x^2 = 25(54^2+x^2)

49x^2 = 25 \times 54^2+ 25x^2

49x^2-25x^2 = 25 \times 54^2

24x^2 = 25 \times 54^2

x^2 = \dfrac{25 \times 54^2}{24}

x =\sqrt{ \dfrac{25 \times 54^2}{24}}

x =\dfrac{5 \times 54}{\sqrt{24}}

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x =\dfrac{45 \times 6}{ 2 \sqrt{ 6}}

x =\dfrac{45 \sqrt{6}}{ 2}

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